Probability and Distributions

Transcription

1 Probability and Distributions What is a statistical model? A statistical model is a set of assumptions by which the hypothetical population distribution of data is inferred. It is typically postulated that the statistical model of density function is determined by parameters, and that data is a realization of iid random variables, called a random sample. Once the statistical model is constructed, basic notion of random variables and their distributions can be readily applicable; thus, the solid understanding of probability theory in terms of calculus (as opposed to measure theory) is the key to a successful investigation of statistical model. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 1 / 44

2 Concept of random variables. A numerically valued function X of an outcome ω from a sample space Ω X : Ω R : ω X (ω) is called a random variable (r.v.), and usually determined by an experiment. We conventionally denote random variables by uppercase letters X, Y, Z, U, V,... from the end of the alphabet. In particular, a discrete random variable is a random variable that can take values on a finite set {a 1, a 2,..., a n } of real numbers (usually integers), or on a countably infinite set {a 1, a 2, a 3,...}. The statement such as X = a i is an event since {ω : X (ω) = a i } is a subset of the sample space Ω. Thus, we can consider the probability of the event {X = a i }, denoted by P(X = a i ). The function p(a i ) := P(X = a i ) over the possible values of X, say a 1, a 2,..., is called a frequency function, or a probability mass function. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 2 / 44

3 Continuous random variable. A continuous random variable is a random variable whose possible values are real values such as 78.6, 5.7, 10.24, and so on. Examples of continuous random variables include temperature, height, diameter of metal cylinder, etc. In what follows, a random variable always means a continuous random variable, unless it is particularly said to be discrete. The probability distribution of a random variable X specifies how its values are distributed over the real numbers. This is completely characterized by the cumulative distribution function (cdf). The cdf F (t) := P(X t). represents the probability that the random variable X is less than or equal to t. Then we often say that the random variable X is distributed as F. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 3 / 44

4 Probability distributions. It is often the case that the probability of the random variable X being in any particular range of values is given by the area under a curve over that range of values. This curve is called the probability density function (pdf) of the random variable X, denoted by f (x). Thus, the probability that a X b can be expressed as P(a X b) = b a f (x)dx. Furthermore, we can find the following relationship between cdf and pdf: F (t) = P(X t) = t f (x)dx. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 4 / 44

5 Joint distributions. When a pair (X, Y ) of random variables is considered, a joint density function f (x, y) is used to compute probabilities constructed from the random variables X and Y simultaneously. Given the joint density function f (x, y), the distribution for each of X and Y is called the marginal distribution. The marginal density functions of X and Y, denoted by f X (x) and f Y (y), are given respectively by f X (x) = f (x, y) dy and f Y (y) = f (x, y) dx. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 5 / 44

6 Conditional probability distributions. Suppose that two random variables X and Y has a joint density function f (x, y). If f X (x) > 0, then we can define the conditional density function f Y X (y x) given X = x by f Y X (y x) = f (x, y) f X (x). Similarly we can define the conditional density function f X Y (x y) given Y = y by f (x, y) f X Y (x y) = f Y (y) if f Y (y) > 0. Then, clearly we have the following relation f (x, y) = f Y X (y x)f X (x) = f X Y (x y)f Y (y). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 6 / 44

7 Independent random variables. If the joint density function f (x, y) for continuous random variables X and Y is expressed in the form f (x, y) = f X (x)f Y (y) for all x, y, then X and Y are said to be independent. Similarly we can construct a joint density function f (x 1,..., x n ) for a vector (X 1,..., X n ) of random variables. If it satisfies f (x 1,..., x n ) = f 1 (x 1 ) f n (x n ) for all x 1,..., x n, then X 1,..., X n are said to be independent. Furthermore, if X 1,..., X n share the common density function f then X 1,..., X n is called independent and identically distributed, or simply, iid for short. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 7 / 44

8 Expectation. Let X be a random variable, and let f is the pdf of X. Then the expectation (or expected value, or mean) of the random variable X is given by E[X ] = x f (x) dx. (1.1) We denote the expected value (1.1) by E(X ), µ, or µ X. For a function g, we define the expectation E[g(X )] of the function g(x ) of the random variable X by E[g(X )] = g(x) f (x) dx. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 8 / 44

9 Expectation, continued. One can think of the expectation E(X ) as an operation on a random variable X which returns the average value for X. It is also useful to observe that this is a linear operator satisfying the following properties: [ ] n n E a + b i X i = a + b i E[X i ] i=1 with random variables X 1,..., X n and constants a and b 1,..., b n. i=1 Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 9 / 44

10 Variance and covariance. The variance of a random variable X, denoted by Var(X ), is the expected value of the squared difference between the random variable and its expected value E(X ). Thus, it is given by Var(X ) := E[(X E(X )) 2 ] = E[X 2 ] (E[X ]) 2. The square-root Var(X ) is called the standard error (SE) (or standard deviation (SD)) of the random variable X. Now suppose that we have two random variables X and Y. Then the covariance of two random variables X and Y can be defined as Cov(X, Y ) := E((X µ x )(Y µ y )) = E(XY ) E(X ) E(Y ), where µ x = E(X ) and µ y = E(Y ). The covariance Cov(X, Y ) measures the strength of the dependence of the two random variables. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 10 / 44

11 Variance and covariance, continued. In contrast to the expectation, the variance is not a linear operator: The variance of ax + b with constants a and b is given by Var(aX + b) = a 2 Var(X ). For two random variables X and Y, we have Var(X + Y ) = Var(X ) + Var(Y ) + 2Cov(X, Y ). (1.2) However, if X and Y are independent, then Cov(X, Y ) = 0, and consequently Var(X + Y ) = Var(X ) + Var(Y ). In general, when we have a sequence of independent random variables X 1,..., X n, we have Var(X X n ) = Var(X 1 ) + + Var(X n ). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 11 / 44

12 Normal distributions. A normal distribution is represented by a family of distributions which have the same general shape, sometimes described as bell shaped. The normal distribution has the pdf f (x) := 1 σ 2π exp [ ] (x µ)2, (1.3) 2σ 2 which depends upon two parameters µ and σ 2. Here π = is the famous pi (the ratio of the circumference of a circle to its diameter), and exp(u) is the exponential function e u with the base e = of the natural logarithm. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 12 / 44

13 Normal distributions, continued. Much celebrated integrations are the following: 1 σ 2π 1 σ 2π 1 σ 2π e (x µ) 2 2σ 2 dx = 1, (1.4) xe (x µ) 2 2σ 2 dx = µ, (x µ) 2 e (x µ) 2 2σ 2 dx = σ 2. The integration (1.4) guarantees that the density function (1.3) always represents a probability density no matter what values the parameters µ and σ 2 are chosen. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 13 / 44

14 Parameters of normal distributions. We say that a random variable X is normally distributed with parameter (µ, σ 2 ) when X has the density function (1.3). The parameter µ, called mean (or, location parameter), provides the center of the density, and the density function f (x) is symmetric around µ. The parameter σ is a standard deviation (or, a scale parameter); small values of σ lead to high peaks but sharp drops. Larger values of σ lead to flatter densities. The shorthand notation X N(µ, σ 2 ) is often used to express that X is a normal random variable with parameter (µ, σ 2 ). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 14 / 44

15 Linear transform of normal random variable. One of the important properties of normal distribution is that if X is a normal random variable with parameter (µ, σ 2 ) [that is, the pdf of X is given by the density function (1.3)]. then Y = ax + b is also a normal random variable having parameter (aµ + b, (aσ) 2 ). In particular, X µ (1.5) σ becomes a normal random variable with parameter (0, 1), called the z-score. Then the normal density φ(x) with parameter (0, 1) becomes φ(x) := 1 2π e x2 2, which is known as the standard normal density. We define the cdf Φ by Φ(t) := t φ(x) dx. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 15 / 44

16 Gamma distribution. The gamma function, denoted by Γ(α), is defined as Γ(α) = Then the gamma density is defined as f (t) = 0 u α 1 e u du, α > 0. λα Γ(α) tα 1 e λt t 0 which depends on two parameters α > 0 and λ > 0. [A statistics textbook often uses a parameter β = 1 λ instead.] Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 16 / 44

17 Parameters of gamma/exponential distribution. We call the parameter α a shape parameter, because changing α changes the shape of the density. We call the parameter λ a rate parameter, because changing λ merely rescales the density without changing its shape. In particular, the gamma distribution with α = 1 becomes an exponential distribution (with parameter λ). The parameter β = 1/λ, if you choose, is called a scale parameter. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 17 / 44

18 Chi-square distribution. The gamma distribution f (x) = 1 2 n/2 Γ(n/2) x n/2 1 e x/2, x 0 with α = n 2 and λ = 1 is called the chi-square distribution with n 2 degrees of freedom. It plays a vital role in understanding another important distribution, called t-distribution. A chi-square random variable X has the mean E[X ] = n and the variance Var(X ) = 2n. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 18 / 44

19 Chi-square distribution of one degree of freedom. Let X be a standard normal random variable, and let Y = X 2 be the square of X. Then we can express the cdf F Y of Y as F Y (t) = P(X 2 t) = P( t X t) = Φ( t) Φ( t) in terms of the standard normal cdf Φ. By differentiating the cdf, we can obtain the pdf f Y as follows. f Y (t) = d dt F Y (t) = 1 2π t 1 2 e t 2 = 1 2 1/2 Γ(1/2) t 1/2 e t/2, where we use Γ(1/2) = π. Thus, the square X 2 of X is the chi-square random variable with 1 degree of freedom. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 19 / 44

20 Chi-square distribution with n df. Let X 1,..., X n be iid standard normal random variables. Since X 2 i s have the gamma distribution with parameter (1/2, 1/2), the sum Y = n i=1 X 2 i has the gamma distribution with parameter (n/2, 1/2). That is, the sum Y has the chi-square distribution with n degree of freedom. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 20 / 44

21 Student s t-distribution. Let X be a chi-square random variable with n degrees of freedom, and let Y be a standard normal random variable. Suppose that X and Y are independent. Then the distribution of the quotient Z = Y X /n is called the Student s t-distribution with n degrees of freedom. The pdf f Z (z) is given by f Z (z) = Γ((n + 1)/2) nπγ(n/2) (1 + z 2 /n) (n+1)/2, < z <. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 21 / 44

22 Sample statistics. Let X 1,..., X n be independent and identically distributed random variables ( iid random variables for short). Then the random variable n X = 1 n i=1 X i is called the sample mean, and the random variable S 2 = 1 n 1 is called the sample variance. n (X i X ) 2 i=1 Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 22 / 44

23 Sample statistics under normal assumption. Assume that X 1,..., X n are iid normal random variables with parameter (µ, σ 2 ). Then the sample statistics X and S 2 have the following properties: 1 E[ X ] = µ and Var( X ) = σ2 n. 2 The sample variance S 2 can be rewritten as ( n ) S 2 = 1 Xi 2 n n 1 X 2 i=1 ( n ) = 1 (X i µ) 2 n( n 1 X µ) 2. i=1 Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 23 / 44

24 Sample statistics under normal assumption, continued. In particular, we can obtain ( n ) E[S 2 ] = 1 E[(X i µ) 2 ] ne[( n 1 X µ) 2 ] i=1 ( n ) = 1 Var(X i ) nvar( n 1 X ) = σ 2. i=1 X has the normal distribution with parameter (µ, σ 2 /n), and X and S 2 are independent. Furthermore, W n 1 = (n 1)S 2 σ 2 = 1 σ 2 n (X i X ) 2 i=1 has the chi-square distribution with (n 1) degrees of freedom. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 24 / 44

25 t-statistic. Assuming iid normal random variables X 1,..., X n with parameter (µ, σ 2 ), we can construct a standard normal random variable Z 1 = X µ σ/ and a chi-square random variable W n n 1 = (n 1)S2. σ 2 Furthermore, the function of random sample ) T = X µ S/ n = ( X µ σ/ n (n 1)S 2 /σ 2 n 1 = Z 1 W n 1 n 1 has the t-distribution with (n 1) degrees of freedom, and it is called t-statistic. The statistic S = S 2 is called the sample standard deviation. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 25 / 44

26 Transformation of one random variable. Let X be a random variable with pdf f X (x), and let g(x) be a differentiable and strictly monotonic function (that is, either strictly increasing or strictly decreasing) on real line. Then the inverse function g 1 exists, and the random variable Y = g(x ) has the pdf f Y (y) = f X (g 1 (y)) d dy g 1 (y). (1.6) Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 26 / 44

27 Linear transform of random variable. Let X be a random variable with pdf f X (x). Then we can define the linear transform Y = ax + b with real values a and b. To compute the density function f Y (y) of the random variable Y, we can apply (1.6) to obtain ( ) y b 1 f Y (y) = f X a a. For example, suppose that X is a normal random variable with parameter (µ, σ 2 ). Then the density function f Y (y) for the linear transform Y = ax + b becomes ] 1 f Y (y) = [ ( a σ) 2π exp (y (aµ + b))2, 2(aσ) 2 which implies that Y is a normal random variable with parameter (aµ + b, (aσ) 2 ). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 27 / 44

28 Change of variables in double integral. For a rectangle A = {(x, y) : a 1 x b 1, a 2 y b 2 } on a plane, the integration of a function f (x, y) over A is formally written as A f (x, y) dx dy = b2 b1 a 2 a 1 f (x, y) dx dy (1.7) Suppose that a transformation (g 1 (x, y), g 2 (x, y)) is differentiable and has the inverse transformation (h 1 (u, v), h 2 (u, v)) satisfying { { x = h 1 (u, v) u = g 1 (x, y) if and only if. (1.8) y = h 2 (u, v) v = g 2 (x, y) Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 28 / 44

29 Change of variables in double integral, continued. Then we can state the change of variable in the double integral (1.7) as follows: (i) define the inverse image of A by h 1 (A) = g(a) = {(g 1 (x, y), g 2 (x, y)) : (x, y) A}, (ii) calculate the Jacobian J h (u, v) = det [ h u 1(u, v) h u 2(u, v) h ] v 1(u, v) h, v 2(u, v) (iii) finally, if it does not vanish for all x, y, then we can obtain f (x, y) dx dy = f (h 1 (u, v), h 2 (u, v)) J h (u, v) du dv. A g(a) Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 29 / 44

30 Transformation of two random variables. Suppose that (a) random variables X and Y has a joint density function f X,Y (x, y), (b) a transformation (g 1 (x, y), g 2 (x, y)) is differentiable, and (c) its inverse transformation (h 1 (u, v), h 2 (u, v)) satisfies (1.8). Then the random variables U = g 1 (X, Y ) and V = g 2 (X, Y ) has the joint density function f U,V (u, v) = f X,Y (h 1 (u, v), h 2 (u, v)) J h (u, v). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 30 / 44

31 Order statistics. Let X 1,..., X n be independent and identically distributed random variables (iid random variables) with the common cdf F (x) and the pdf f (x). Then we can define the random variable V = min(x 1,..., X n ) of the minimum of X 1,..., X n. To find the distribution of V, consider the survival function P(V > x) of V and calculate as follows. P(V > x) = P(X 1 > x) P(X n > x) = [1 F (x)] n. Thus, we can obtain the cdf F V (x) and the pdf f V (x) of V F V (x) = 1 [1 F (x)] n and f V (x) = d dx F V (x) = nf (x)[1 F (x)] n 1. In particular, if X 1,..., X n be independent exponential random variables with the common parameter λ, then the minimum min(x 1,..., X n ) has the density (nλ)e (nλ)x which is again the exponential density function with parameter (nλ). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 31 / 44

32 Order statistics,continued. Let X 1,..., X n be iid random variables with the common cdf F (x) and the pdf f (x). When we sort X 1,..., X n as X (1) < X (2) < < X (n), the random variable X (k) is called the k-th order statistic. Theorem The pdf of the k-th order statistic X (k) is given by f k (x) = n! (k 1)!(n k)! f (x)f k 1 (x)[1 F (x)] n k. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 32 / 44

33 Uniform and beta distribution. When X 1,..., X n be iid uniform random variables on [0, θ], the pdf of X (k) /θ becomes f k (x) = n! (k 1)!(n k)! x k 1 (1 x) n k, 0 x 1, which is known as the beta distribution with parameters α = k and β = n k + 1. Uniform on [0, θ] Beta(k, n f (x) = 1 θ, 0 x θ E[X ] = θ/2, Var(X ) = θ 2 /12 X (k) /θ Γ(n + 1) f (x) = Γ(k)Γ(n + 1 k) E[X ] = k n + 1, Var Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 33 / 44

34 Summary of discrete random variables. 1 X is the number of Bernoulli trials until the first success occurs. 2 X is the number of Bernoulli trials until the rth success occurs. If X 1,, X r are independent geometric random variables with parameter p, then X = r k=1 X k becomes a negative binomial random variable with parameter (r, p). 3 r = 1. 4 If X 1,, X n are independent Bernoulli trials, then X = n k=1 X k becomes a binomial random variable with (n, p). 5 If X 1 and X 2 are independent binomial random variables with (n 1, p) and (n 2, p) then so is X = X 1 + X 2 with (n 1 + n 2, p). 6 Binomial distribution converges to Poisson distribution by letting n and p 0 while λ = np. 7 If X 1 and X 2 are independent Poisson random variables with λ 1 and λ 2 then so is X = X 1 + X 2 with λ 1 + λ 2. Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 34 / 44

35 Summary of discrete random variables, continued. Bernoulli trial (a) P(X = 1) = p and P(X = 0) = q := 1 p E[X ] = p, Var(X ) = pq The probability p of success and the probability q := 1 p of failure Geometric(p) P(X = i) = q i 1 p, i = 1, 2,... E[X ] = 1 p, Var(X ) = q p 2 B ( P(X = i) = E[X ] = Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 35 / 44

36 Summary of continuous random variables. 1 If X 1 and X 2 are independent normal random variables with (µ 1, σ 2 1) and (µ 2, σ 2 2) then so is X = X 1 + X 2 with (µ 1 + µ 2, σ σ 2 2). 2 If X 1,, X m are iid normal random variables with parameter (µ, σ 2 ), then X = m ) 2 k=1 is a chi-square random variable with m df. 3 n = m 2 and λ = 1 2. ( Xk µ σ 4 If X 1,..., X n are independent exponential random variables with λ 1,, λ n then so is X = min{x 1,..., X n } with n i=1 λ i. 5 If X 1,, X n are iid exponential random variables with parameter λ, then X = n k=1 X k is a gamma random variable with (n, λ). 6 n = 1. 7 If X 1 and X 2 are independent gamma random variables with parameters (n 1, λ) and (n 2, λ) then so is X = X 1 + X 2 with (n 1 + n 2, λ). Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 36 / 44

37 Summary of continuous random variables, continued. (d Normal(µ, σ 2 ) Ex f (x) = 1 2πσ e (x µ)2 2σ 2, < x < E[X ] = µ, Var(X ) = σ 2 f (x) E[X ] = (a) (b) (c) (e) G Note #1 Mathematical Statistics/August 30, 2018 Probability and Distributions 37 / 44 λe

38 Exercises Note #1 Mathematical Statistics/August 30, 2018 Exercises 38 / 44

39 Problem Suppose that X has the pdf f (x) = (x + 2)/18, 2 x 4. Calculate E[X ], E[(X + 2) 3 ], and E[6X 2(X + 2) 3 ]. Note #1 Mathematical Statistics/August 30, 2018 Exercises 39 / 44

40 Problem Suppose that X has the pdf f (x) = (x + 2)/18, 2 x 4. Calculate E[X ], E[(X + 2) 3 ], and E[6X 2(X + 2) 3 ]. 4 [ ] (x + 2) 2 (x + 2) 3 4 E[X + 2] = dx = = 4; thus, E[X ] = [ ] E[(X + 2) 3 (x + 2) 4 (x + 2) 5 4 ] = dx = = 432/ E[6X 2(X + 2) 3 ] = 6E[X ] 2E[(X + 2) 3 ] = 804/5. Note #1 Mathematical Statistics/August 30, 2018 Exercises 39 / 44

41 Problem Let f (x) = 6x(1 x), 0 x 1, be a pdf for X. Find the mean and the variance. Note #1 Mathematical Statistics/August 30, 2018 Exercises 40 / 44

42 Problem Let f (x) = 6x(1 x), 0 x 1, be a pdf for X. Find the mean and the variance. 1 E[X ] = (6x 2 6x 3 )dx = [2x 3 3x ] 4 1 = 1/ [ 3x E[X 2 ] = (6x 3 6x 4 4 )dx = 0 2 6x ] 5 1 = 3/ Therefore, Var(X ) = 3/10 (1/2) 2 = 1/20. Note #1 Mathematical Statistics/August 30, 2018 Exercises 40 / 44

43 Problem 1 If X N(75, 100), calculate P(X 60) and P(70 X 100). 2 If X N(µ, σ 2 ), find b so that P( b < (X µ)/σ < b) = Note #1 Mathematical Statistics/August 30, 2018 Exercises 41 / 44

44 Problem 1 If X N(75, 100), calculate P(X 60) and P(70 X 100). 2 If X N(µ, σ 2 ), find b so that P( b < (X µ)/σ < b) = P(X 60) = Φ ( ) P(70 X 100) = Φ ( ) ( Φ ) b Note #1 Mathematical Statistics/August 30, 2018 Exercises 41 / 44

45 Problem Suppose that X and Y have the joint density f (x, y) = 2e x y, 0 < x < y <. Find the joint density for U = 2X and V = Y X. Note #1 Mathematical Statistics/August 30, 2018 Exercises 42 / 44

46 Problem Suppose that X and Y have the joint density f (x, y) = 2e x y, 0 < x < y <. Find the joint density for U = 2X and V = Y X. The transformation g 1 (x, y) = 2x and g 2 (x, y) = y x maps from {(x, y) : 0 < x < y < } to {(u, v) : 0 < u <, 0 < v < }. Then we can obtain the inverse transformation x = h 1 (u, v) = u/2 and y = h 2 (u, v) = u/2 + v, and calculate the Jacobian [ ] 1/2 0 J h (u, v) = det = 1/2. 1/2 1 Thus, the joint distribution is given by f UV (u, v) = f XY (u/2, u/2 + v)/2 = e u v if 0 < u < and 0 < v <. Note #1 Mathematical Statistics/August 30, 2018 Exercises 42 / 44

47 Problem Suppose that X and Y have the joint density f (x, y) = 8xy, 0 < x < y < 1. Find the joint density for U = X /Y and V = Y. Note #1 Mathematical Statistics/August 30, 2018 Exercises 43 / 44

48 Problem Suppose that X and Y have the joint density f (x, y) = 8xy, 0 < x < y < 1. Find the joint density for U = X /Y and V = Y. The transformation g 1 (x, y) = x/y and g 2 (x, y) = y maps from {(x, y) : 0 < x < y < 1} to {(u, v) : 0 < u < 1, 0 < v < 1}. Then we can obtain the inverse transformation x = h 1 (u, v) = uv and y = h 2 (u, v) = v, and calculate the Jacobian [ ] v u J h (u, v) = det = v. 0 1 Thus, the joint distribution is given by if 0 < u < 1 and 0 < v < 1. f UV (u, v) = f XY (uv, v) v = 8uv 3 Note #1 Mathematical Statistics/August 30, 2018 Exercises 43 / 44

49 Problem Let X (k), k = 1, 2, 3, 4, be the order statistics of a random sample of size 4 from the pdf f (x) = e x, 0 < x <. Find P(X (4) > 3). Note #1 Mathematical Statistics/August 30, 2018 Exercises 44 / 44

50 Problem Let X (k), k = 1, 2, 3, 4, be the order statistics of a random sample of size 4 from the pdf f (x) = e x, 0 < x <. Find P(X (4) > 3). P(X (4) 3) = 4 P(X i 3) = (1 e 3 ) 4. Thus, we obtain i=1 P(X (4) > 3) = 1 (1 e 3 ) 4. Note #1 Mathematical Statistics/August 30, 2018 Exercises 44 / 44